The Cubic regression spline with a knot at $\xi$ has the form $$ f(x) = \beta_{0} + \beta_{1}x + \beta_{2}x^{2} + \beta_{3}x^{3} + \beta_{4}\left(x-\xi\right)^{3}_{+} $$
Find a choice of $a_{1}$, $b_{1}$, $c_{1}$ and $d_{1}$ for,
$$ f_{1}(x) = a_{1} + b_{1}x + c_{1}x^{2} + d_{1}x^{3} $$
for $x<\xi$. Equating the two equations,
$$ a_{1} + b_{1}x + c_{1}x^{2} + d_{1}x^{3}= \beta_{0} + \beta_{1}x + \beta_{2}x^{2} + \beta_{3}x^{3} $$
and equating coefficients we have,
$$ a_{1} = \beta_{0} \qquad b_{1} = \beta_{1} \\c_{1} = \beta_{2} \qquad d_{1} = \beta_{3}. \\$$
Find a choice of $a_{2}$, $b_{2}$, $c_{2}$ and $d_{2}$ for,
$$ f_{2}(x) = a_{2} + b_{2}x + c_{2}x^{2} + d_{2}x^{3} $$
for $x \geq \xi$. Equating the two equations,
$$ a_{2} + b_{2}x + c_{2}x^{2} + d_{2}x^{3}= \beta_{0} + \beta_{1}x + \beta_{2}x^{2} + \beta_{3}x^{3} + \beta_{4}\left(x-\xi\right)^{3} $$
Expanding the cubic term,
$$ \left(x-\xi\right)^{3} = x^3 -3\xi x^2 + 3\xi^{2} x - \xi^3. $$
Substituting the expanded expression in the equation and comparing coefficients gives,
$$ a_{2} = \beta_{0} - \beta_{4}\xi^{3} \qquad b_{2} = \beta_{1} + 3\beta{4}\xi^{2} \\c_{2} = \beta_{2} - 3\beta_{4}\xi \qquad d_{2} = \beta_{3} + \beta_{4}. \\$$
Show that $f(x)$ is continuous at the knot.
$$ f_{1}(\xi) = \beta_{0} + \beta_{1}\xi + \beta_{2}\xi^{2} + \beta_{3}\xi^{3} $$
and
$$ \begin{aligned} f_{2}(\xi) &= \beta_{0} - \beta_{4}\xi^{3} + \beta_{1}\xi + 3\beta_{4}\xi^{3} + \beta_{2}\xi^{2} - 3\beta_{4}\xi^{3} + \beta_{3}\xi^{3} + \beta_{4}\xi^{3} \\&= \beta_{0} + \beta_{1}\xi + \beta_{2}\xi^{2} + \beta_{3}\xi^{3} + \left(-\xi^{3} +3\xi^{3} -3\xi^{3} + \xi^{3}\right)\beta_{4} \\&= \beta_{0} + \beta_{1}\xi + \beta_{2}\xi^{2} + \beta_{3}\xi^{3} \\&= f_{1}(\xi) \end{aligned} $$
Therefore $f(x)$ is continous at the knot $\xi$ (and everywhere else as $f_{1}$ and $f_{2}$ are continuous functions).
Show that $f'(x)$ is continuous at the knot. $$ f'{1}(\xi) = \beta + 2\beta_{2}\xi + 3\beta_{3}\xi^{2} $$
and
$$ \begin{aligned} f'{2}(\xi) &= \beta + 3\beta_{4}\xi^{2} + 2\beta_{2}\xi - 6\beta_{4}\xi^{2} + 3\beta_{3}\xi^{2} + 3\beta_{4}\xi^{2} \\&= \beta_{1} + 2\beta_{2}\xi + 3\beta_{3}\xi^{2} + \left(3\xi^{2} -6\xi^{2} +3\xi^{2} \right)\beta_{4} \\&= \beta_{1} + 2\beta_{2}\xi + 3\beta_{3}\xi^{2} \\&= f'_{1}(\xi) \end{aligned} $$
Therefore the first derivative of $f$ is continuous at the knot and everywhere.
Show that $f''(x)$ is continuous at the knot. $$ f''{1}(\xi) = 2\beta + 6\beta_{3}\xi $$
and
$$ \begin{aligned} f''{2}(\xi) &= 2\beta - 6\beta_{4}\xi + 6\beta_{3}\xi + 6\beta_{4}\xi \\&= 2\beta_{2} + 6\beta_{3}\xi + \left(-6\xi + 6\xi \right)\beta_{4} \\&= 2\beta_{2} + 6\beta_{3}\xi \\&= f''_{1}(\xi) \end{aligned} $$
Therefore the second derivative of $f$ is continuous at the knot and everywhere.
Considering the two smoothing functions, $$ g_1 = \text{arg} \min_{g}\left(\sum_{i}\left(y_i - g_i(x)\right)^2 + \lambda\int\left(g^{(3)}(t)\right)dt\right) \\g_2 = \text{arg} \min_{g}\left(\sum_{i}\left(y_i - g_i(x)\right)^2 + \lambda\int\left(g^{(4)}(t)\right)dt\right) $$
For $\lambda \rightarrow \infty$ the effect of the constraint in maximized. In this case it reduces the amount of roughness of a function to the differential order of the constrained term. As $g_1$ has a lower order of smoothness constrained, it will have the lower training error than $g_2$.
It is not possible to definitively say which will have a greater test RSS as that depends on the variance and bias tradeoff. All we can say is that maximally constraining the $g_1$ will lower the variance and increase the bias of the model than maximally constraining $g_2$ will.
For $\lambda = 0$ there is no constraint and the training data is completely interpolated by the function and most definitely overfits the data and has a poor test RSS. Here $g_1 = g_2$.